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3 years ago

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Ok we trying this again. so just in case if the picture isnt showing, here are the questions:

1 answer:

11Alexandr11 [23.1K]3 years ago

3 0

Answer:

1. 31.68 moles of water, H₂O

2. 14.81 moles of Cr

Explanation:

1. Determination of the number of mole of water, H₂O.

The balanced equation for the reaction is given below:

C₁₂H₂₂O₁₁ —> 12C + 11H₂O

From the balanced equation above,

1 mole of C₁₂H₂₂O₁₁ produced 12 moles of C and 11 moles of H₂O.

Next, we shall determine the number of mole C₁₂H₂₂O₁₁ needed to produce 34.55 moles of C. This can be obtained as follow:

From the balanced equation above,

1 mole of C₁₂H₂₂O₁₁ produced 12 moles of C.

Therefore, Xmol of C₁₂H₂₂O₁₁ will produce 34.55 moles of C i.e

Xmol of C₁₂H₂₂O₁₁ = 34.55 / 12

Xmol of C₁₂H₂₂O₁₁ = 2.88 moles

Thus, 2.88 moles of C₁₂H₂₂O₁₁ is needed.

Finally, we shall determine the number of mole of water, H₂O produced from the reaction. This can be obtained as follow:

From the balanced equation above,

1 mole of C₁₂H₂₂O₁₁ produced 11 moles of H₂O.

Therefore, 2.88 moles of C₁₂H₂₂O₁₁ will produce = 2.88 × 11 = 31.68 moles of H₂O.

Thus, 31.68 moles of water, H₂O were obtained from the reaction.

2. Determination of the number of mole of Cr needed.

The balanced equation for the reaction is given below:

Cr + H₂SO₄ —> CrSO₄ + H₂

From the balanced equation above,

1 mole of Cr reacted to produce 1 mole of CrSO₄.

Finally, we shall determine the number of mole of Cr needed to produce 14.81 moles of CrSO₄. This can be obtained as follow:

From the balanced equation above,

1 mole of Cr reacted to produce 1 mole of CrSO₄.

Therefore, 14.81 moles of Cr will also react to produce 14.81 moles of CrSO₄.

Thus, 14.81 moles of Cr is needed for the reaction.

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Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

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$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

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$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

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x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

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